Question

Objects of masses m1=4.00kg and m2 = 9.00kg are connected by a light string that passes over a frictionless pulley. The object m1 is held at rest on the floor, and m2 rests on a fixed incline of (theta)=40.0 degree. The objects are released from rest, and m2 slides 1.00m down the incline in 4.00s. Determine
a) the acceleration of each object
b) the tension in the string, and
c) the coefficient of kinetic friction between m2 and he incline.

Please help!
Thank You!

Answer 1

Since M1 and M2 are connected together the acceleration is the same in both cases and can be f=determined using the formula \[x=(at^{2})/2\] or \[a=(2x)/t^{2}\]This gives the acceleration for both masses.

The net force (F net) on M2 is given by the following equation\[F_{net_{2}} =F_{x}-T -\mu N\] Since \[F_{x}=M_{2}g \sin(\theta) \]\[F_{g_{2}}=M_{2}g\] and \[ \mu N= \mu M_{2}g \cos (\theta)\]Then the equation for the net force can be written as \[F_{net_{2}}=M_{2}a= M_{2}g \sin(\theta)-T - \mu M_{2}g \cos (\theta)\]There are 2 unknowns in the above equation (the frictional coeffiecent mu) and the tension (T)

The net force on M1 is given by \[F_{net_{1}} =T-F_{g_{1}}\] where \[F_{g_{1}}=M_{1}g\] therefore \[F_{net_{1}}=M_{1}a= T-M_{1}g \] We can determine the tension in the string by solving for T. That is, \[ T=M_{1}a +M_{1}g\]
By making mu the argument we can now substitute the value for T and obtain the value for the frictional coefficent such that \[ \mu = (M_{2}g \sin(\theta)-M_{2}a-T)/(M_{2}g \cos (\theta))\]